Calculus Examples

$1 + \frac{1}{x} - \frac{1}{x^{2}}$

Step 1

Write $1 + \frac{1}{x} - \frac{1}{x^{2}}$ as a function.

$f (x) = 1 + \frac{1}{x} - \frac{1}{x^{2}}$

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Differentiate.

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Step 2.1.1.1

By the Sum Rule, the derivative of $1 + \frac{1}{x} - \frac{1}{x^{2}}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [- \frac{1}{x^{2}}]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [- \frac{1}{x^{2}}]$

Step 2.1.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [- \frac{1}{x^{2}}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [\frac{1}{x}]$ .

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Step 2.1.2.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$0 + \frac{d}{d x} [x^{- 1}] + \frac{d}{d x} [- \frac{1}{x^{2}}]$

Step 2.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 1$ .

$0 - x^{- 2} + \frac{d}{d x} [- \frac{1}{x^{2}}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- \frac{1}{x^{2}}]$ .

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Step 2.1.3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x^{2}}$ .

$0 - x^{- 2} - \frac{d}{d x} [\frac{1}{x^{2}}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 2.1.3.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$0 - x^{- 2} - \frac{d}{d x} [{(x^{2})}^{- 1}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 2.1.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 2.1.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$0 - x^{- 2} - (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 2.1.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 1$ .

$0 - x^{- 2} - (- u^{- 2} \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 2.1.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$0 - x^{- 2} - (- {(x^{2})}^{- 2} \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 2.1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$0 - x^{- 2} - (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 2.1.3.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$0 - x^{- 2} - (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.6

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 2.1.3.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$0 - x^{- 2} - (- x^{2 \cdot - 2} (2 x)) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.6.2

Multiply $2$ by $- 2$ .

$0 - x^{- 2} - (- x^{- 4} (2 x)) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.7

Multiply $2$ by $- 1$ .

$0 - x^{- 2} - (- 2 x^{- 4} x) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.8

Raise $x$ to the power of $1$ .

$0 - x^{- 2} - (- 2 (x^{1} x^{- 4})) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.9

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$0 - x^{- 2} - (- 2 x^{1 - 4}) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.10

Subtract $4$ from $1$ .

$0 - x^{- 2} - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.11

Multiply $- 2$ by $- 1$ .

$0 - x^{- 2} + 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0$

Step 2.1.3.12

Multiply $\frac{1}{x^{2}}$ by $0$ .

$0 - x^{- 2} + 2 x^{- 3} + 0$

Step 2.1.3.13

Add $2 x^{- 3}$ and $0$ .

$0 - x^{- 2} + 2 x^{- 3}$

Step 2.1.4

Simplify.

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Step 2.1.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$0 - \frac{1}{x^{2}} + 2 x^{- 3}$

Step 2.1.4.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$0 - \frac{1}{x^{2}} + 2 \frac{1}{x^{3}}$

Step 2.1.4.3

Combine terms.

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Step 2.1.4.3.1

Subtract $\frac{1}{x^{2}}$ from $0$ .

$- \frac{1}{x^{2}} + 2 \frac{1}{x^{3}}$

Step 2.1.4.3.2

Combine $2$ and $\frac{1}{x^{3}}$ .

$f' (x) = - \frac{1}{x^{2}} + \frac{2}{x^{3}}$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $- \frac{1}{x^{2}} + \frac{2}{x^{3}}$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [\frac{2}{x^{3}}]$ .

$\frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [- \frac{1}{x^{2}}]$ .

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Step 2.2.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x^{2}}$ .

$- \frac{d}{d x} [\frac{1}{x^{2}}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$- \frac{d}{d x} [{(x^{2})}^{- 1}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 2.2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $x^{2}$ .

$- (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = - 1$ .

$- (- {u_{1}}^{- 2} \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.3.3

Replace all occurrences of $u_{1}$ with $x^{2}$ .

$- (- {(x^{2})}^{- 2} \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$- (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.6

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 2.2.2.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- (- x^{2 \cdot - 2} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.6.2

Multiply $2$ by $- 2$ .

$- (- x^{- 4} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.7

Multiply $2$ by $- 1$ .

$- (- 2 x^{- 4} x) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.8

Raise $x$ to the power of $1$ .

$- (- 2 (x^{1} x^{- 4})) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.9

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- (- 2 x^{1 - 4}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.10

Subtract $4$ from $1$ .

$- (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.11

Multiply $- 2$ by $- 1$ .

$2 x^{- 3} + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.12

Multiply $\frac{1}{x^{2}}$ by $0$ .

$2 x^{- 3} + 0 + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.2.13

Add $2 x^{- 3}$ and $0$ .

$2 x^{- 3} + \frac{d}{d x} [\frac{2}{x^{3}}]$

Step 2.2.3

Evaluate $\frac{d}{d x} [\frac{2}{x^{3}}]$ .

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Step 2.2.3.1

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{x^{3}}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{x^{3}}]$ .

$2 x^{- 3} + 2 \frac{d}{d x} [\frac{1}{x^{3}}]$

Step 2.2.3.2

Rewrite $\frac{1}{x^{3}}$ as ${(x^{3})}^{- 1}$ .

$2 x^{- 3} + 2 \frac{d}{d x} [{(x^{3})}^{- 1}]$

Step 2.2.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{3}$ .

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Step 2.2.3.3.1

To apply the Chain Rule, set $u_{2}$ as $x^{3}$ .

$2 x^{- 3} + 2 (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d x} [x^{3}])$

Step 2.2.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 1$ .

$2 x^{- 3} + 2 (- {u_{2}}^{- 2} \frac{d}{d x} [x^{3}])$

Step 2.2.3.3.3

Replace all occurrences of $u_{2}$ with $x^{3}$ .

$2 x^{- 3} + 2 (- {(x^{3})}^{- 2} \frac{d}{d x} [x^{3}])$

Step 2.2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$2 x^{- 3} + 2 (- {(x^{3})}^{- 2} (3 x^{2}))$

Step 2.2.3.5

Multiply the exponents in ${(x^{3})}^{- 2}$ .

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Step 2.2.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2 x^{- 3} + 2 (- x^{3 \cdot - 2} (3 x^{2}))$

Step 2.2.3.5.2

Multiply $3$ by $- 2$ .

$2 x^{- 3} + 2 (- x^{- 6} (3 x^{2}))$

Step 2.2.3.6

Multiply $3$ by $- 1$ .

$2 x^{- 3} + 2 (- 3 x^{- 6} x^{2})$

Step 2.2.3.7

Multiply $x^{- 6}$ by $x^{2}$ by adding the exponents.

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Step 2.2.3.7.1

Move $x^{2}$ .

$2 x^{- 3} + 2 (- 3 (x^{2} x^{- 6}))$

Step 2.2.3.7.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 x^{- 3} + 2 (- 3 x^{2 - 6})$

Step 2.2.3.7.3

Subtract $6$ from $2$ .

$2 x^{- 3} + 2 (- 3 x^{- 4})$

Step 2.2.3.8

Multiply $- 3$ by $2$ .

$2 x^{- 3} - 6 x^{- 4}$

Step 2.2.4

Simplify.

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Step 2.2.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 \frac{1}{x^{3}} - 6 x^{- 4}$

Step 2.2.4.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 \frac{1}{x^{3}} - 6 \frac{1}{x^{4}}$

Step 2.2.4.3

Combine terms.

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Step 2.2.4.3.1

Combine $2$ and $\frac{1}{x^{3}}$ .

$\frac{2}{x^{3}} - 6 \frac{1}{x^{4}}$

Step 2.2.4.3.2

Combine $- 6$ and $\frac{1}{x^{4}}$ .

$\frac{2}{x^{3}} + \frac{- 6}{x^{4}}$

Step 2.2.4.3.3

Move the negative in front of the fraction.

$f'' (x) = \frac{2}{x^{3}} - \frac{6}{x^{4}}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2}{x^{3}} - \frac{6}{x^{4}}$ .

$\frac{2}{x^{3}} - \frac{6}{x^{4}}$

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{2}{x^{3}} - \frac{6}{x^{4}} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$\frac{2}{x^{3}} - \frac{6}{x^{4}} = 0$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x^{3}, x^{4}, 1$

Step 3.2.2

Since $x^{3}, x^{4}, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1$ then find LCM for the variable part $x^{3}, x^{4}$ .

Step 3.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 3.2.5

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 3.2.6

The factors for $x^{3}$ are $x \cdot x \cdot x$ , which is $x$ multiplied by each other $3$ times.

$x^{3} = x \cdot x \cdot x$

$x$ occurs $3$ times.

Step 3.2.7

The factors for $x^{4}$ are $x \cdot x \cdot x \cdot x$ , which is $x$ multiplied by each other $4$ times.

$x^{4} = x \cdot x \cdot x \cdot x$

$x$ occurs $4$ times.

Step 3.2.8

The LCM of $x^{3}, x^{4}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x \cdot x \cdot x$

Step 3.2.9

Simplify $x \cdot x \cdot x \cdot x$ .

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Step 3.2.9.1

Multiply $x$ by $x$ .

$x^{2} \cdot x \cdot x$

Step 3.2.9.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 3.2.9.2.1

Multiply $x^{2}$ by $x$ .

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Step 3.2.9.2.1.1

Raise $x$ to the power of $1$ .

$x^{2} \cdot x^{1} \cdot x$

Step 3.2.9.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{2 + 1} \cdot x$

Step 3.2.9.2.2

Add $2$ and $1$ .

$x^{3} \cdot x$

Step 3.2.9.3

Multiply $x^{3}$ by $x$ by adding the exponents.

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Step 3.2.9.3.1

Multiply $x^{3}$ by $x$ .

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Step 3.2.9.3.1.1

Raise $x$ to the power of $1$ .

$x^{3} \cdot x^{1}$

Step 3.2.9.3.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{3 + 1}$

Step 3.2.9.3.2

Add $3$ and $1$ .

$x^{4}$

Step 3.3

Multiply each term in $\frac{2}{x^{3}} - \frac{6}{x^{4}} = 0$ by $x^{4}$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{2}{x^{3}} - \frac{6}{x^{4}} = 0$ by $x^{4}$ .

$\frac{2}{x^{3}} x^{4} - \frac{6}{x^{4}} x^{4} = 0 x^{4}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Cancel the common factor of $x^{3}$ .

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Step 3.3.2.1.1.1

Factor $x^{3}$ out of $x^{4}$ .

$\frac{2}{x^{3}} (x^{3} x) - \frac{6}{x^{4}} x^{4} = 0 x^{4}$

Step 3.3.2.1.1.2

Cancel the common factor.

$\frac{2}{x^{3}} (x^{3} x) - \frac{6}{x^{4}} x^{4} = 0 x^{4}$

Step 3.3.2.1.1.3

Rewrite the expression.

$2 x - \frac{6}{x^{4}} x^{4} = 0 x^{4}$

Step 3.3.2.1.2

Cancel the common factor of $x^{4}$ .

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Step 3.3.2.1.2.1

Move the leading negative in $- \frac{6}{x^{4}}$ into the numerator.

$2 x + \frac{- 6}{x^{4}} x^{4} = 0 x^{4}$

Step 3.3.2.1.2.2

Cancel the common factor.

$2 x + \frac{- 6}{x^{4}} x^{4} = 0 x^{4}$

Step 3.3.2.1.2.3

Rewrite the expression.

$2 x - 6 = 0 x^{4}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Multiply $0$ by $x^{4}$ .

$2 x - 6 = 0$

Step 3.4

Solve the equation.

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Step 3.4.1

Add $6$ to both sides of the equation.

$2 x = 6$

Step 3.4.2

Divide each term in $2 x = 6$ by $2$ and simplify.

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Step 3.4.2.1

Divide each term in $2 x = 6$ by $2$ .

$\frac{2 x}{2} = \frac{6}{2}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $2$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{6}{2}$

Step 3.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{6}{2}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

Divide $6$ by $2$ .

$x = 3$

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

Substitute $3$ in $f (x) = 1 + \frac{1}{x} - \frac{1}{x^{2}}$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = 1 + \frac{1}{3} - \frac{1}{{(3)}^{2}}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Raise $3$ to the power of $2$ .

$f (3) = 1 + \frac{1}{3} - \frac{1}{9}$

Step 4.1.2.2

Find the common denominator.

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Step 4.1.2.2.1

Write $1$ as a fraction with denominator $1$ .

$f (3) = \frac{1}{1} + \frac{1}{3} - \frac{1}{9}$

Step 4.1.2.2.2

Multiply $\frac{1}{1}$ by $\frac{9}{9}$ .

$f (3) = \frac{1}{1} \cdot \frac{9}{9} + \frac{1}{3} - \frac{1}{9}$

Step 4.1.2.2.3

Multiply $\frac{1}{1}$ by $\frac{9}{9}$ .

$f (3) = \frac{9}{9} + \frac{1}{3} - \frac{1}{9}$

Step 4.1.2.2.4

Multiply $\frac{1}{3}$ by $\frac{3}{3}$ .

$f (3) = \frac{9}{9} + \frac{1}{3} \cdot \frac{3}{3} - \frac{1}{9}$

Step 4.1.2.2.5

Multiply $\frac{1}{3}$ by $\frac{3}{3}$ .

$f (3) = \frac{9}{9} + \frac{3}{3 \cdot 3} - \frac{1}{9}$

Step 4.1.2.2.6

Multiply $3$ by $3$ .

$f (3) = \frac{9}{9} + \frac{3}{9} - \frac{1}{9}$

Step 4.1.2.3

Combine the numerators over the common denominator.

$f (3) = \frac{9 + 3 - 1}{9}$

Step 4.1.2.4

Simplify by adding and subtracting.

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Step 4.1.2.4.1

Add $9$ and $3$ .

$f (3) = \frac{12 - 1}{9}$

Step 4.1.2.4.2

Subtract $1$ from $12$ .

$f (3) = \frac{11}{9}$

Step 4.1.2.5

The final answer is $\frac{11}{9}$ .

$\frac{11}{9}$

Step 4.2

The point found by substituting $3$ in $f (x) = 1 + \frac{1}{x} - \frac{1}{x^{2}}$ is $(3, \frac{11}{9})$ . This point can be an inflection point.

$(3, \frac{11}{9})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, 3) \cup (3, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 3)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $2.9$ in the expression.

$f'' (2.9) = \frac{2}{{(2.9)}^{3}} - \frac{6}{{(2.9)}^{4}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Raise $2.9$ to the power of $3$ .

$f'' (2.9) = \frac{2}{24.389} - \frac{6}{{(2.9)}^{4}}$

Step 6.2.1.2

Divide $2$ by $24.389$ .

$f'' (2.9) = 0.08200418 - \frac{6}{{(2.9)}^{4}}$

Step 6.2.1.3

Raise $2.9$ to the power of $4$ .

$f'' (2.9) = 0.08200418 - \frac{6}{70.7281}$

Step 6.2.1.4

Divide $6$ by $70.7281$ .

$f'' (2.9) = 0.08200418 - 1 \cdot 0.08483191$

Step 6.2.1.5

Multiply $- 1$ by $0.08483191$ .

$f'' (2.9) = 0.08200418 - 0.08483191$

Step 6.2.2

Subtract $0.08483191$ from $0.08200418$ .

$f'' (2.9) = - 0.00282773$

Step 6.2.3

The final answer is $- 0.00282773$ .

$- 0.00282773$

Step 6.3

At $2.9$ , the second derivative is $- 0.00282773$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 3)$

Decreasing on $(- \infty, 3)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(3, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $3.1$ in the expression.

$f'' (3.1) = \frac{2}{{(3.1)}^{3}} - \frac{6}{{(3.1)}^{4}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $3.1$ to the power of $3$ .

$f'' (3.1) = \frac{2}{29.791} - \frac{6}{{(3.1)}^{4}}$

Step 7.2.1.2

Divide $2$ by $29.791$ .

$f'' (3.1) = 0.06713436 - \frac{6}{{(3.1)}^{4}}$

Step 7.2.1.3

Raise $3.1$ to the power of $4$ .

$f'' (3.1) = 0.06713436 - \frac{6}{92.3521}$

Step 7.2.1.4

Divide $6$ by $92.3521$ .

$f'' (3.1) = 0.06713436 - 1 \cdot 0.06496874$

Step 7.2.1.5

Multiply $- 1$ by $0.06496874$ .

$f'' (3.1) = 0.06713436 - 0.06496874$

Step 7.2.2

Subtract $0.06496874$ from $0.06713436$ .

$f'' (3.1) = 0.00216562$

Step 7.2.3

The final answer is $0.00216562$ .

$0.00216562$

Step 7.3

At $3.1$ , the second derivative is $0.00216562$ . Since this is positive, the second derivative is increasing on the interval $(3, \infty)$ .

Increasing on $(3, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(3, \frac{11}{9})$ .

$(3, \frac{11}{9})$

Step 9